Question
The random variable $X$ can take only the values $0,1,2,3$. Given that $P(X=0)=P(X=1)=p$ and $P(X=2)=$ $P ( X =3)$ such that $\Sigma p_i x_i^2=2 \Sigma p_i x_i$, find the value of p .
✓
Answer
Given $P(X=0)=P(X=1)=p$ and $P(X=2)=P(X=3)=k($ say $)$
The probability distribution of the random variable X is
We know that $\sum p _{ i }=1$
$
\begin{array}{l}
\Rightarrow p+p+k+k=1 \Rightarrow 2 p+2 k=1 \\
\Rightarrow p+k=\frac{1}{2} \Rightarrow k=\frac{1}{2}-p
\end{array}
$
We construct the following table:
$
\begin{array}{l}
\text { Given } \Sigma p_i x_i^2=2 \Sigma p_i x_i \\
\Rightarrow \frac{13}{2}-12 p=2\left(\frac{5}{2}-4 p\right) \Rightarrow \frac{13}{2}-12 p=5-8 p \Rightarrow-4 p=5-\frac{13}{2} \\
\Rightarrow-4 p=-\frac{3}{2} \Rightarrow p=\frac{3}{8}
\end{array}
$
Hence, the value of $p$ is $\frac{3}{8}$.
The probability distribution of the random variable X is
| X | 0 | 1 | 2 | 3 |
| P(X) | p | p | k | k |
$
\begin{array}{l}
\Rightarrow p+p+k+k=1 \Rightarrow 2 p+2 k=1 \\
\Rightarrow p+k=\frac{1}{2} \Rightarrow k=\frac{1}{2}-p
\end{array}
$
We construct the following table:
| xi | Pi | Pixi | Pix2i |
| 0 | p | 0 | 0 |
| 1 | p | p | p |
| 2 | $\frac{1}{2}=p$ | 1-2p | 2-4p |
| 3 | $\frac{1}{2}=p$ | $\frac{3}{2}-3 p$ | $\frac{9}{2}-9 p$ |
| Total | $\frac{5}{2}=4 p$ | $\frac{13}{2}-12 p$ |
\begin{array}{l}
\text { Given } \Sigma p_i x_i^2=2 \Sigma p_i x_i \\
\Rightarrow \frac{13}{2}-12 p=2\left(\frac{5}{2}-4 p\right) \Rightarrow \frac{13}{2}-12 p=5-8 p \Rightarrow-4 p=5-\frac{13}{2} \\
\Rightarrow-4 p=-\frac{3}{2} \Rightarrow p=\frac{3}{8}
\end{array}
$
Hence, the value of $p$ is $\frac{3}{8}$.
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